Giải pt: \(\dfrac{x-5}{x+6}-\dfrac{2x^2+23x+61}{x^2+x-30}=\dfrac{x+6}{5-x}\)
Giải các phương trình sau:
a) \(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)
b) \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)
a, đk : x khác 5;-6
\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm)
b, đk : x khác 1;3
\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)
pt vô nghiệm
a, đk : x khác 5;-6
x2+12x+36+x2−10x+25=2x2+23x+61x2+12x+36+x2−10x+25=2x2+23x+61
⇔2x+61=23x+61⇔21x=0⇔x=0⇔2x+61=23x+61⇔21x=0⇔x=0(tm)
b, đk : x khác 1;3
x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)
pt vô nghiệm
a: \(\Leftrightarrow\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)
\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
=>x=0(nhận)
b: \(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)
\(\Leftrightarrow x^2+2x-15=x^2-1-8\)
=>2x-15=-9
=>2x=-6
hay x=-3(nhận)
\(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2=x-30}\)
ĐKXĐ: x<>5; x<>-6
\(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)
\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
=>23x+61=2x+61
=>x=0
giải phương trình :
a) \(\dfrac{x+2}{x+1}+\dfrac{3}{x-2}=\dfrac{3}{x^2-x-2}+1\)
b)\(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)
a) điều kiện xác định : \(x\ne2;x\ne-1\)
ta có : \(\dfrac{x+2}{x+1}+\dfrac{3}{x-2}=\dfrac{3}{x^2-x-2}+1\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-2\right)+3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3+x^2-x-2}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow x^2-4+3x+3=x^2-x+1\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\left(tmđk\right)\)
vậy \(x=\dfrac{1}{2}\)
b) điều kiện xác định : \(x\ne5;x\ne-6\)
ta có : \(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)
\(\Leftrightarrow\dfrac{\left(x+6\right)^2+\left(x-5\right)^2}{\left(x-5\right)\left(x+6\right)}=\dfrac{2x^2+23x+61}{\left(x-5\right)\left(x+6\right)}\)
\(\Rightarrow x^2+12x+36+x^2-25x+25=2x^2+23x+61\)
\(\Leftrightarrow-13x=23x\Leftrightarrow x=0\left(tmđk\right)\)
vậy \(x=0\)
giải pt
\(\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\)
\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)
`1/(3-x)-1/(x+1)=x/(x-3)-(x-1)^2/(x^2-2x-3)(x ne -1,3)`
`<=>(-x-1)/(x^2-2x-3)-(x-3)/(x^2-2x-3)=(x^2+x)/(x^2-2x-3)-(x-1)^2/(x^2-2x-3)`
`<=>-x-1-x+3=x^2+x-x^2+2x-1`
`<=>-2x+2=3x-1`
`<=>5x=3`
`<=>x=3/5`
Vậy `S={3/5}`
`1/(x-2)-6/(x+3)=6/(6-x^2-x)(x ne 2,-3)`
`<=>(x+3)/(x^2+x-6)-(6x-12)/(x^2+x-6)+6/(x^2+x-6)=0`
`<=>x+3-6x+12+6=0`
`<=>-5x+21=0`
`<=>x=21/5`
Vậy `S={21/5}`
a) ĐKXĐ: \(x\notin\left\{3;-1\right\}\)
Ta có: \(\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\)
\(\Leftrightarrow\dfrac{-1\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}-\dfrac{x-3}{\left(x+1\right)\left(x-3\right)}=\dfrac{x\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}-\dfrac{x^2-2x+1}{\left(x-3\right)\left(x+1\right)}\)
Suy ra: \(-x-1-x+3=x^2+x-x^2+2x-1\)
\(\Leftrightarrow3x-1=-2x+2\)
\(\Leftrightarrow3x+2x=2+1\)
\(\Leftrightarrow5x=3\)
hay \(x=\dfrac{3}{5}\)(nhận)
Vậy: \(S=\left\{\dfrac{3}{5}\right\}\)
Bài 1:
a) Giải PT sau: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
b) Giải PT sau: |2x+6|-x=3
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
GIẢI PT:
a) \(\dfrac{x}{x-5}=\dfrac{x-2}{x-6}\)
b) \(\dfrac{2x}{8-x}-\dfrac{2-2x}{4-x}=1\)
e) \(\dfrac{2x}{x+4}-\dfrac{4x}{x^2-16}=0\)
MN GIẢI BÀI NÀY GIÚP E VỚI Ạ. E ĐANG CẦN GẤP Ạ.
\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
bài tập 1: giải các phương trình sau:
a, \(\dfrac{3}{1-4x}\)= \(\dfrac{2}{4x+1}\)- \(\dfrac{8+6x}{16x^2-1}\)
b, \(\dfrac{3}{5x-1}\)+\(\dfrac{2}{3-5x}\)=\(\dfrac{4}{\left(1-5x\right)\left(5x-3\right)}\)
c, \(\dfrac{x+2}{x+1}\)+\(\dfrac{3}{x-2}\)=\(\dfrac{3}{x^2-x-2}+1\)
d, \(\dfrac{5-x}{4x^2-8x}+\dfrac{7}{8}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8x-16}\)
e, \(\dfrac{x+6}{x+5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)
f, \(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)
các bn giúp mik với!! vài câu cx được
a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)
=>-12x-4=2x-10
=>-14x=-6
hay x=3/7
b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)
=>15x-9-10x+2=-4
=>5x-7=-4
=>5x=3
hay x=3/5(loại)
c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\)
=>4x=2
hay x=1/2(nhận)
giải pt sau
a)\(\dfrac{60}{x}=\dfrac{4}{3}+\dfrac{60-x}{x+4}\)
b)\(\dfrac{100}{x}-\dfrac{100}{x+20}=\dfrac{5}{6}\)
c)\(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
Helppppp
b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)
=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)
=>x*(x+20)=400*6=2400
=>x^2+20x-2400=0
=>(x+60)(x-40)=0
=>x=-60 hoặc x=40
c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
=>(2x+1)^2-(2x-1)^2=8
=>4x^2+4x+1-4x^2+4x-1=8
=>8x=8
=>x=1(nhận)
Giải PT sau:
a, 3x - 7 = 0
b, 8 - 5x = 0
c, 3x - 2 = 5x + 8
d, \(\dfrac{3x-2}{3}\) = \(\dfrac{1-x}{2}\)
e, ( 5x + 1)(x - 3) = 0
f, (x + 1)(2x - 3) = 0
g, 4x(x + 3) - 5(x + 3) = 0
h, 8(x - 6) - 2x(6 - x) = 0
i, \(\dfrac{2}{x-1}\) + \(\dfrac{1}{x}\) = \(\dfrac{2x+5}{x^2-x}\)
k, \(\dfrac{3}{x+2}\) - \(\dfrac{2}{x-2}\) = \(\dfrac{2-x}{x^2-4}\)
m, \(\dfrac{3}{x}\) - \(\dfrac{2}{x-3}\) = \(\dfrac{4-x}{x^2-3}\)
n,\(\dfrac{3}{2x+10}\)+ \(\dfrac{2x}{x^2-25}\) = \(\dfrac{3}{x-5}\)
u, \(\dfrac{2}{x+3}\) - \(\dfrac{3}{x-2}\) = \(\dfrac{x+4}{\left(x+3\right)\left(x-2\right)}\)
a, 3x - 7 = 0
<=> 3x = 7
<=> x = 7/3
b, 8 - 5x = 0
<=> -5x = -8
<=> x = 8/5
c, 3x - 2 = 5x + 8
<=> -2x = 10
<=> x = -5
e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)
`a ) 3x - 7 = 0`
`\(\Leftrightarrow \) 3x = 7`
`\(\Leftrightarrow \) x = 7/3`
Vậy `S = {-7/3}`
Giải pt:
a) \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x(x^4+4x^2+16)}\)
b) \(\dfrac{4x^{2} + 10}{x^{2} + 5} - \dfrac{9}{x^{2} + 4} = \dfrac{8}{x^{2} + 3} + \dfrac{6}{x^{2} + 1}\)
điều kiện xác định \(x\ne0\)
ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-\left(x^3+2x^2+4x-2x^2-4x-8\right)}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-x^3-2x^2-4x+2x^2+4x+8}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{-x^2+2x+12}{x^4+4x^2+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)
\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)
tới đây bn bấm máy tính nha